Intuition behind Taylor/Maclaurin Series

Question

** This is a different question than Intuition explanation of taylor expansion? **

I understand some of the intuition behind a Taylor/Maclaurin expansion. More specifically, I understand that adding higher and higher degree polynomials will add more 'turning points' on a graph to better represent the curves of the function you wish to approximate.

I don't understand why

a.) you add the terms; shouldn't adding terms shift the graph left/right, up/down? In addition to the question of shifting the graph, I just don't understand why you would add more terms, rather than just change your first term accordingly.

I now understand the above, thanks to microarm15 and Nicholas Stull. I now just do not understand part b of this question

b.) the terms added are the successive derivatives of the function. What does adding successive derivatives mean/give you?

Any help on the matter is greatly appreciated. Thanks!

Answer 1

Well, first of all I am not a mathematician, and sorry if I cannot explain what I know in a plain way.

Most popular functions, i.e. like trigonometric, log, exp, are non linear functions, which means the higher the order of the approximating polynomial, the better the approximation. Increasing the order of the polynomial's degree means increasing the non linearity; and yes, that should shift the value of the function a little bit with adding every higher term towards the real value of the function, which is mostly transcendental, i.e. cannot be finitely represented.

Note: in real life, non linear functions, specially the popular transcendetal fuctions like trigonometric which are extensively used in GPUs, are approximated by chebychev polynomials which provide much higher accuracy, with the same degree, than taylor polynomials.

Answer 2

b.) the terms added are the successive derivatives of the function. What does adding successive derivatives mean/give you?

Any help on the matter is greatly appreciated. Thanks!

Look at the $N$th order Taylor polynomial for $f(x)$ at $x=a$: $$T_N(x):=\sum_{n=0}^N {f^{(n)}(a)\over n!}(x-a)^n.$$

You say "adding successive derivatives of the function". Well, not exactly. What is happening is that as we add more terms in the Taylor polynomial for $f$, we are adding on higher and higher powers of $x$. The coefficients of those particular terms are related to $f$ via the formula above. But we are not adding on "derivatives of $f$", rather higher order polynomial terms.

The reason this is helpful is because if you think back to how the $N$th degree Taylor polynomial was derived: $T_N(x)$ was the polynomial of degree $N$ that matched up through the first $N$ derivatives of $f$ at $x=a$. That is, $${d^j\over dx^j}T_N(x)\Bigg|_{x=a}={d^j\over dx^j}f(x)\Bigg|_{x=a}\forall j=0,1,\dots,N$$.

So for example, $T_1(x)$ is the best linear approximation of $f$ at $x=a$ in the sense that $T_1(x)$ matches $f$ in function value and derivative at $x=a$.

Similarly, $T_2(x)$ is the "best quadratic approximation" of $f$ at $x=a$. And $T_3(x)$ is the best cubic approximation of $f$ at $x=a$. Geometrically, as we had more and more terms to the Taylor polynomial, the result is a better and better approximation (in the sense described above).

Here's a picture of $T_N(x)$, $N=0,\dots,12$ for $f(x)=\sin x$ at $a=0$:

Finally, to connect this back to Taylor series, just let $N\to\infty$. You might find this post helpful.